Question: Simplify
\[\frac{\tan 30^\circ + \tan 40^\circ + \tan 50^\circ + \tan 60^\circ}{\cos 20^\circ}.\]
Answer: In general, from the angle addition formula,
\begin{align*}
\tan x + \tan y &= \frac{\sin x}{\cos x} + \frac{\sin y}{\cos y} \\
&= \frac{\sin x \cos y + \sin y \cos x}{\cos x \cos y} \\
&= \frac{\sin (x + y)}{\cos x \cos y}.
\end{align*}Thus,
\begin{align*}
\frac{\tan 30^\circ + \tan 40^\circ + \tan 50^\circ + \tan 60^\circ}{\cos 20^\circ} &= \frac{\frac{\sin 70^\circ}{\cos 30^\circ \cos 40^\circ} + \frac{\sin 110^\circ}{\cos 50^\circ \cos 60^\circ}}{\cos 20^\circ} \\
&= \frac{1}{\cos 30^\circ \cos 40^\circ} + \frac{1}{\cos 50^\circ \cos 60^\circ} \\
&= \frac{2}{\sqrt{3} \cos 40^\circ} + \frac{2}{\cos 50^\circ} \\
&= 2 \cdot \frac{\cos 50^\circ + \sqrt{3} \cos 40^\circ}{\sqrt{3} \cos 40^\circ \cos 50^\circ} \\
&= 4 \cdot \frac{\frac{1}{2} \cos 50^\circ + \frac{\sqrt{3}}{2} \cos 40^\circ}{\sqrt{3} \cos 40^\circ \cos 50^\circ} \\
&= 4 \cdot \frac{\cos 60^\circ \sin 40^\circ + \sin 60^\circ \cos 40^\circ}{\sqrt{3} \cos 40^\circ \cos 50^\circ}.
\end{align*}From the angle addition formula and product-to-sum formula,
\begin{align*}
4 \cdot \frac{\cos 60^\circ \sin 40^\circ + \sin 60^\circ \cos 40^\circ}{\sqrt{3} \cos 40^\circ \cos 50^\circ} &= 4 \cdot \frac{\sin (60^\circ + 40^\circ)}{\sqrt{3} \cdot \frac{1}{2} (\cos 90^\circ + \cos 10^\circ)} \\
&= \frac{8 \sin 100^\circ}{\sqrt{3} \cos 10^\circ} \\
&= \frac{8 \cos 10^\circ}{\sqrt{3} \cos 10^\circ} \\
&= \boxed{\frac{8 \sqrt{3}}{3}}.
\end{align*}